# ncert solutions for class 8 maths chapter 2 linear equations in one variable in English | class 8 chapter 2 maths exercise 2.1, 2.2, 2.3

Students can practice the questions based on the same through class 8th maths ncert solutions for class 8 maths chapter 2 exercise 2.1, ncert solutions for class 8 maths chapter 2 exercise 2.2, ncert solutions for class 8 maths chapter 2 exercise 2.3 and many more. These examples are in the form of word problems.

The next topic is Solving equations having the variable on both sides. In this section, students will learn how to solve such equations which have expressions with the variable on both sides. class 8th maths

Here we solve ncert solutions for ncert maths class 8 chapter 2 solution From class 8 maths chapter 2 pdf concepts all solutions with easy method with expert solutions. chapter 2 maths class 8 help students in their study, home work and preparing for exam. Soon we provide CBSE chapter 2 maths class 8 solutions English Medium. NCERT solutions for class 8 maths chapter 2 pdf Linear Equations in One Variable PDF here. You can download ncert english book class 8 from official NCERT website or** Click HERE**.

# ncert solutions for class 8 maths chapter 2 linear equations in one variable in English

class 8th maths

## ncert solutions for class 8 maths chapter 2 Linear Equations in One Variable in English

### NCERT solutions for class 8 maths chapter 2 pdf┬а– Questions Answer Click Here

## ncert solutions for class 8 maths chapter 2 EXERCISE 2.1 class 8th maths

**Solve the following equations.**

1. ЁЭСе тАУ 2 = 7 | 2. ЁЭСж + 3 = 10 | 3. 6 = ЁЭСз + 2 |

4. \(\displaystyle \frac{3}{7}+x=\frac{{17}}{7}\) | 5. 6ЁЭСе = 12 | 6. \(\displaystyle \frac{t}{5}=10\) |

7. \(\displaystyle \frac{{2x}}{3}=18\) | 8. \(\displaystyle 1.6=\frac{y}{{1.5}}\) | 9. 7ЁЭСе тАУ 9 = 16 |

10. 14ЁЭСж тАУ 8 = 13 | 11. 17 + 6ЁЭСЭ = 9 | 12. \(\displaystyle \frac{x}{3}+1=\frac{7}{{15}}\) |

Solutions:

**1. ЁЭСе тАУ 2 = 7**

Sol.: ЁЭСе – 2 = 7

ЁЭСе = 7 + 2 (by transposing)

ЁЭСе = 9 answer

**2. ЁЭСж + 3 = 10**

Sol.:┬а ЁЭСж + 3 = 10

ЁЭСж = 10 – 3 (by transposing)

ЁЭСж = 7 answer

**3. 6 = ЁЭСз + 2**

Sol.: 6 = ЁЭСз + 2

or ЁЭСз + 2 = 6

ЁЭСз = 6 – 2 *(by transposing)*

ЁЭСз = 4 answer

**4. \(\displaystyle \frac{3}{7}+x=\frac{{17}}{7}\)**

Sol.: \(\displaystyle \frac{3}{7}+x=\frac{{17}}{7}\)

\(\displaystyle x=\frac{{17}}{7}-\frac{3}{7}\) *(by transposing)*\(\displaystyle x=\frac{{17-3}}{7}\)

*(by L.C.M)*

\(\displaystyle x=\frac{{14}}{7}\)

ЁЭСе = 2 answer

**6. 6ЁЭСе = 12**

Sol.: 6ЁЭСе = 12

\(\displaystyle x=\frac{{12}}{6}\)

ЁЭСе = 2 answer

**7. \(\displaystyle \frac{{2x}}{3}=18\)**

Sol.: \(\displaystyle \frac{{2x}}{3}=18\)

\(\displaystyle \frac{{2x}}{3}=\frac{{18}}{1}\)

2ЁЭСе ├Ч 1 = 18 ├Ч 3 *(by cross multiplication)*2ЁЭСе = 54

\(\displaystyle \frac{{54}}{2}\)ЁЭСе = 27

*answer*

**8. \(\displaystyle 1.6=\frac{y}{{1.5}}\)**

Sol.: \(\displaystyle 1.6=\frac{y}{{1.5}}\)\(\displaystyle \frac{{1.6}}{1}=\frac{y}{{1.5}}\)ЁЭСж = 1.5 ├Ч 1.6

ЁЭСж = 2.4 answer

**9. 7ЁЭСе тАУ 9 = 16**

Sol.: 7ЁЭСе – 9 = 16

7ЁЭСе = 16 + 9

7ЁЭСе = 25

\(\displaystyle \frac{{25}}{7}\) answer

**10. 14ЁЭСж тАУ 8 = 13**

Sol.: 14ЁЭСж – 8 = 13

14ЁЭСж = 13 + 8

14ЁЭСж = 21

ЁЭСж = \(\displaystyle \frac{{21}}{{14}}\)

ЁЭСж = \(\displaystyle \frac{3}{2}\) answer

**11. 17 + 6ЁЭСЭ = 9**

Sol.: 17 + 6ЁЭСЭ = 9

6ЁЭСЭ = 9 – 17 (by transposing)

6ЁЭСЭ = – 8

\(\displaystyle \frac{{-8}}{6}\)

\(\displaystyle \frac{{-4}}{3}\) answer

**12. \(\displaystyle \frac{x}{3}+1=\frac{7}{{15}}\)**

Sol.: \(\displaystyle \frac{x}{3}+1=\frac{7}{{15}}\)

\(\displaystyle \frac{x}{3}=\frac{7}{{15}}-\frac{1}{1}\) (by transposing)

\(\displaystyle \frac{x}{3}=\frac{{7-1}}{{15}}\) (by L.C.M.)

\(\displaystyle \frac{x}{3}=\frac{6}{{15}}\)

15 ├Ч ЁЭСе = 6 ├Ч 3 (by cross multiplication)

15ЁЭСе = 18

\(\displaystyle x=\frac{{18}}{{15}}\)

\(\displaystyle x=\frac{{6}}{{5}}\) answer

## Ncert Solutions for class 8 maths chapter 2 EXERCISE 2.2

class 8th maths

**1. If you subtract \(\displaystyle \frac{1}{2}\) from a number and multiply the result by \(\displaystyle \frac{1}{2}\), you get \(\displaystyle \frac{1}{8}\). What is the number?**

Sol.: Let’s the number is = ЁЭСе

subtract \(\displaystyle \frac{1}{2}\) from x = \(\displaystyle x-\frac{1}{2}\)

Result is \(\displaystyle \frac{{2x-1}}{2}\)

Result multiply by \(\displaystyle \frac{1}{2}\)

According to the question –

\(\displaystyle \begin{array}{l}\left( {\frac{{2x-1}}{2}} \right)\times \frac{1}{2}=\frac{1}{8}\\\frac{{2x-1}}{4}=\frac{1}{8}\end{array}\)

\(\displaystyle 8\times (2x-1)=4\times 1\) (by cross multifaction)

16ЁЭСе – 8 = 4┬а

16ЁЭСе = 4 + 8 (by transposing)

16ЁЭСе = 12

\(\displaystyle \begin{array}{l}x=\frac{{12}}{{16}}\\x=\frac{3}{4}\end{array}\)

so that number is \(\displaystyle \frac{3}{4}\) answer.

**2. The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool?**

Sol.: Let’s the breadth of the swimming pool is = ЁЭСе┬а

so Its length is = 2ЁЭСе + 2

Perimeter of a rectangle is = 154 m

2 (length + breadth) = 154

put the value –

2 (2ЁЭСе + 2 + ЁЭСе) = 154

2 (3ЁЭСе +2) = 154

6ЁЭСе + 4 = 154

6ЁЭСе = 154 – 4

6ЁЭСе = 150

\(\displaystyle x=\frac{{150}}{6}\)

ЁЭСе = 25

so the breadth of the pool is ЁЭСе = 25 m

and the length of the pool is 2ЁЭСе + 2┬а

2 ├Ч 25 + 2 = 52 m

**3. The base of an isosceles triangle is \(\displaystyle \frac{4}{3}\) cm. The perimeter of the triangle is \(\displaystyle 4\frac{2}{{15}}\) cm. What is the length of either of the remaining equal sides?**

Sol.: Let’s the equal side is = ЁЭСе

Base of isosceles triangle is = \(\displaystyle \frac{4}{3}\) m

Perimeter of the triangle = \(\displaystyle 4\frac{2}{{15}}\)

Sum of all three sides = \(\displaystyle 4\frac{2}{{15}}\)

\(\displaystyle \begin{array}{l}x+x+\frac{4}{3}=4\frac{2}{{15}}\\2x+\frac{4}{3}=\frac{{62}}{{15}}\end{array}\)

\(\displaystyle 2x=\frac{{62}}{{15}}-\frac{4}{3}\) (by transposing)

\(\displaystyle \begin{array}{l}2x=\frac{{62-20}}{{15}}\\2x=\frac{{42}}{{15}}\\2x=\frac{{14}}{5}\\x=\frac{{14}}{5}\times \frac{1}{2}\\x=\frac{7}{5}\end{array}\)

The length of each of the remaining equal sides x = \(\displaystyle \frac{7}{5}\) cm

or \(\displaystyle 1\frac{2}{5}\) cm

**4. Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.**

Sol.: Let’s the first number is = ЁЭСе

and the second number is = ЁЭСе + 15

according to the question –

ЁЭСе + ЁЭСе + 15 = 95

2ЁЭСе + 15 = 95

2ЁЭСе = 95 – 15 (by transposing)

2ЁЭСе = 80

ЁЭСе = \(\displaystyle \frac{{80}}{2}\)

ЁЭСе = 40

so the first number ЁЭСе = 40 answer

and the second number is = ЁЭСе + 15

= 40 + 15

= 55 answer

**5. Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?**

Sol.: Let’s first number is = 5ЁЭСе

and the second number is = 3ЁЭСе

according to the question –

5ЁЭСе- 3ЁЭСе = 18

2ЁЭСе = 18

ЁЭСе = \(\displaystyle \frac{{18}}{2}\)

ЁЭСе = 9

so the first number is 5ЁЭСе = 5 ├Ч 9

= 45 answer

and second number is 3ЁЭСе = 3 ├Ч 9

= 27 answer

**6. Three consecutive integers add up to 51. What are these integers?**

Sol.: Let’s first integers is = ЁЭСе

second integers is = ЁЭСе + 1 (difference of two consecutive integers)

so third integers is = ЁЭСе + 2

according to the question –

ЁЭСе + ЁЭСе + 1 + ЁЭСе + 2 = 51

3ЁЭСе + 3 = 51

3ЁЭСе = 51 – 3 (by transposing)

3ЁЭСе = 48

ЁЭСе = \(\displaystyle \frac{{48}}{3}\)

ЁЭСе = 16

so the first integers ЁЭСе = 16 answer

second integers is ЁЭСе + 1 = 16 + 1 = 17 answer

third integers is ЁЭСе + 2 = 16 + 2 = 18 answer

These three consecutive integers are 16, 17 and 18.

**7. The sum of three consecutive multiples of 8 is 888. Find the multiples.**

Sol.: Let’s first multiples of 8 is = ЁЭСе

second multiple of 8 is = ЁЭСе + 8 (difference of two consecutive multiples of 8)

and third multiple of 8 is = ЁЭСе + 16

according to the question –

ЁЭСе + ЁЭСе + 8 + ЁЭСе + 16 = 888

3ЁЭСе + 24 = 888

3ЁЭСе = 888 – 24

3ЁЭСе = 864

ЁЭСе = \(\displaystyle \frac{{888}}{3}\)

ЁЭСе = 288

So the first multiple of 8 is ЁЭСе = 288

second multiple of 8 is ЁЭСе + 8 = 288 + 8 = 296

third multiple of 8 is ЁЭСе + 16 = 288 + 16 = 304

The three consecutive multiples of 8 is 288, 296 and 304.

**8. Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.**

Sol.: Let’s first integers is = ЁЭСе

Second integers is = ЁЭСе + 1

and third integers is = ЁЭСе + 2

According to the question –

Respectively multiplied by 2, 3 and 4.

2 ├Ч ЁЭСе + 3 ├Ч (ЁЭСе + 1) + 4 ├Ч (ЁЭСе + 2) = 74

2ЁЭСе + 3ЁЭСе + 3 + 4ЁЭСе + 8 = 74

9ЁЭСе + 11 = 74

9ЁЭСе = 74 – 11 (by transposing)

9ЁЭСе = 63

ЁЭСе = \(\displaystyle \frac{{63}}{9}\)

ЁЭСе = 7

So first integers is ЁЭСе = 7

Second integers is ЁЭСе + 1 = 7 + 1 = 8

and third integers is ЁЭСе + 2 =┬а 7 + 2 = 9

These three consecutive integers are 7. 8 and 9.

**9. The ages of Rahul and Haroon are in the ratio 5 : 7. Four years later the sum of their ages will be 56 years. What are their present ages?**

Sol.: Let’s Rahul age is = 5ЁЭСе

and Haroon age is = 7ЁЭСе

After 4 years Rahul age = 5ЁЭСе + 4

and Haroon age = 7ЁЭСе + 4

according to the question –

5ЁЭСе + 4 + 7ЁЭСе + 4 = 56

12ЁЭСе + 8 = 56

12ЁЭСе = 56 – 8 (by transposing)

12ЁЭСе = 48

ЁЭСе = \(\displaystyle \frac{{48}}{12}\)

ЁЭСе = 4

So, Rahul age is 5ЁЭСе = 5 ├Ч 4 = 20 year

and Haroon age is 7ЁЭСе = 7 ├Ч 4 = 28 year┬а

**10. The number of boys and girls in a class are in the ratio 7 : 5. The number of boys is 8 more than the number of girls. What is the total class strength?**

Sol.: Let’s the number of girls are = ЁЭСе

and the number of boys are = ЁЭСе + 8

according to the question –

\(\displaystyle \frac{{x+8}}{x}=\frac{7}{5}\)

\(\displaystyle 7\times x=5\times (x+8)\) (by cross multiplication)

\(\displaystyle \begin{array}{l}7x=5x+40\\7x-5x=40\\2x=40\end{array}\)

ЁЭСе = \(\displaystyle \frac{{40}}{2}\)

ЁЭСе = 20

So the number of girls are ЁЭСе = 20

and the number of boys are ЁЭСе + 8 = 20 + 8 = 28

the total class strength is 20 + 28 = 48 answer

**11. BaichungтАЩs father is 26 years younger than BaichungтАЩs grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?**

Sol.: Let’s Baichung father’s age is = ЁЭСе

Baichung grandfather’s age = ЁЭСе + 26

and Baichung’s age = ЁЭСе – 29

according to the question –

ЁЭСе + ЁЭСе + 26 + ЁЭСе – 29 = 135

3ЁЭСе – 3 = 135

3ЁЭСе = 135 + 3 (by transposing)

3ЁЭСе = 138

ЁЭСе = \(\displaystyle \frac{{138}}{3}\)

ЁЭСе = 46

So, BaichungтАЩs father age ЁЭСе = 46 years

Baichung’s grandfather age ЁЭСе + 26

= 46 + 26 = 72 years

Baichung’s age ЁЭСе – 29

= 46 – 29 = 17 years

**12. Fifteen years from now RaviтАЩs age will be four times his present age. What is RaviтАЩs present age?**

Sol.: Let’s Ravi’s present age is = ЁЭСе

after 15 years Ravi’s age = ЁЭСе + 15

according to the question –

ЁЭСе + 15 = 4ЁЭСе

ЁЭСе – 4x = – 15 (by transposing)

– 3ЁЭСе = – 15

ЁЭСе = \(\displaystyle \frac{{-15}}{-3}\)

ЁЭСе = 5

So Ravi’s present age is = 5 years

**13. A rational number is such that when you multiply it by \(\displaystyle \frac{5}{2}\) and add \(\displaystyle \frac{2}{3}\) to the product, you get \(\displaystyle -\frac{7}{{12}}\). What is the number?**

Sol.: Let’s the rational number is = ЁЭСе

x is multiply by \(\displaystyle \frac{5}{2}\) then product is

\(\displaystyle x\times \frac{5}{2}\)

\(\displaystyle \frac{5x}{2}\)

adding \(\displaystyle \frac{2}{3}\) then

according to the question –

\(\displaystyle \frac{{5x}}{2}+\frac{2}{3}=-\frac{7}{{12}}\)

\(\displaystyle \frac{{15x+4}}{6}=-\frac{7}{{12}}\) (by L.C.M)

\(\displaystyle 12\times (15x+4)=-7\times 6\) (by cross multiplication)

\(\displaystyle \begin{array}{l}180x+48=-42\\180x=-42-48\\180x=-90\\x=\frac{{-90}}{{180}}\end{array}\)

\(\displaystyle x=\frac{{-1}}{2}\)

So rational number is \(\displaystyle \frac{1}{2}\) answer

**14. Lakshmi is a cashier in a bank. She has currency notes of denominations тВ╣ 100, тВ╣ 50 and тВ╣ 10, respectively. The ratio of the number of these notes is 2 : 3 : 5. The total cash with Lakshmi is 4,00,000. How many notes of each denomination does she have?**

Sol.: тИ╡ Ratio of тВ╣ 100 notes is = 2

Let’s number of тВ╣ 100 notes = 2ЁЭСе

So, Value of тВ╣ 100 notes = 2ЁЭСе ├Ч 100 = 200ЁЭСе┬а

тИ╡ Ratio of тВ╣ 50 notes is = 3

Let’s number of тВ╣ 50 notes = 3ЁЭСе

So, Value of тВ╣ 50 notes = 3ЁЭСе ├Ч 50 = 150ЁЭСе

тИ╡ Ratio of тВ╣ 10 notes is = 5

Let’s number of тВ╣ 10 notes = 5ЁЭСе

So, Value of тВ╣ 10 notes = 5ЁЭСе ├Ч 10 = 50ЁЭСе┬а

according to the question –┬а

200ЁЭСе + 150ЁЭСе + 50ЁЭСе = 400000

400ЁЭСе = 400000

\(\displaystyle x=\frac{{400000}}{{400}}\)

ЁЭСе = 1000

So the number of тВ╣100 notes = 2ЁЭСе = 2 ├Ч 1000 = 2000 notes

number of тВ╣ 50 notes = 3ЁЭСе = 3 ├Ч 1000 = 3000 notes

number of тВ╣ 10 notes = 5ЁЭСе = 5 ├Ч 1000 = 5000 notes

**15. I have a total of тВ╣ 300 in coins of denomination тВ╣ 1, тВ╣ 2 and тВ╣ 5. The number of тВ╣ 2 coins is 3 times the number of тВ╣ 5 coins. The total number of coins is 160. How many coins of each denomination are with me?**

Sol.: Let’s number of coins тВ╣ 5 is = *x*

and number of coins тВ╣ 2 is = 3x

Total number of coins is = 160

Then number of coins тВ╣ 1 is = (160 – 3x – x) = (160-4x)

Value of тВ╣ 1 coins = (160 – 4x) ├Ч 1 = 160 – 4x

Value of тВ╣ 2 coins = 3x ├Ч 2 = 6x

Value of тВ╣ 5 coins = x ├Ч 5 = 5x

According to the question –

160 – 4x + 6x + 5x = 300

160 – 4x + 11x = 300

7x = 300 – 160 (by transposing)

7x = 140

\(\displaystyle x=\frac{{140}}{{7}}\)

x = 20

So, Number of тВ╣ 5 coins is = x

= 20 coins

Number of тВ╣ 2 coins is = 3x

= 3 ├Ч 20

= 60 coins

Number of тВ╣ 1 coins is = 160 – 4x

= 160 – 4 ├Ч 20

= 160 – 80┬а

= 80 coins

**16. The organisers of an essay competition decide that a winner in the competition gets a prize of тВ╣ 100 and a participant who does not win gets a prize of тВ╣ 25. The total prize money distributed is тВ╣ 3,000. Find the number of winners, if the total number of participants is 63.**

Sol.: Total number of participants is = 63

Let’s number of winner is = x

and number of participant who does not win = 63 – x

Each winner gets a prize of = тВ╣ 100

So total winner prize money = x ├Ч 100

= 100x

Same, each participant who does not win gets a prize of = тВ╣ 25

So total each participant who does not win gets a prize money = (63 – x) ├Ч 25

= 1575 – 25x

According to the question –

100x + 1575 – 25x = 3000

75x = 3000 – 1575 (by transposing)

75x = 1425

\(\displaystyle x=\frac{{1575}}{{75}}\)

x = 19

So the number of winners is 19 answer.

## Ncert Solutions for Class 8 Maths Chapter 2 EXERCISE 2.3

class 8th maths

**Solve the following equations and check your results.**

1. 3x = 2x + 18 | 2. 5t тАУ 3 = 3t тАУ 5 | 3. 5x + 9 = 5 + 3x |

4. 4z + 3 = 6 + 2z | 5. 2x тАУ 1 = 14 тАУ x | 6. 8x + 4 = 3 (x тАУ 1) + 7 |

7. \(\displaystyle x=\frac{4}{5}(x+10)\) | 8. \(\displaystyle \frac{{2x}}{3}+1=\frac{{7x}}{{15}}+3\) | 9. \(\displaystyle 2y+\frac{5}{3}=\frac{{26}}{3}-y\) |

10. \(\displaystyle 3m=5m-\frac{8}{5}\) | ┬а | ┬а |

**Solutions :**

**1. 3x = 2x + 18**

Sol.: 3x = 2x + 18

3x – 2x = 18 (by transposing)

x = 18 answer┬а

**2. 5t тАУ 3 = 3t тАУ 5**

Sol.: 5t – 3 = 3t – 5

5t – 3t = – 5 + 3

2t = – 2

\(\displaystyle t=-\frac{2}{2}\)

t = – 1 answer┬а

**3. 5x + 9 = 5 + 3x**

Sol.: 5x + 9 = 5 + 3x

5x – 3x = 5 – 9 (by transposing)

2x = – 4

\(\displaystyle x=-\frac{4}{2}\)

x = – 2 answer┬а

**4. 4z + 3 = 6 + 2z**

Sol.: 4z + 3 = 6 + 2z

4z – 2z = 6 – 3 (by transposing)

2z = 3

\(\displaystyle z=\frac{{3}}{{2}}\) answer

**5. 2x тАУ 1 = 14 тАУ x**

Sol.: 2x – 1 = 14 – x

2x + x = 14 + 1 (by transposing)

3x = 15

\(\displaystyle x=\frac{{15}}{{3}}\)

x = 5

**6. 8x + 4 = 3 (x тАУ 1) + 7**

Sol.: 8x + 4 = 3 (x – 1) + 7

8x + 4 = 3x – 3 + 7

8x – 3x = – 3 + 7 –┬а 4 (by transposing)

5x = 7 – 7

5x = 0

\(\displaystyle x=\frac{{0}}{{5}}\)

x = 0 answer

**7. \(\displaystyle x=\frac{4}{5}(x+10)\)**

Sol.: \(\displaystyle x=\frac{4}{5}(x+10)\)

\(\displaystyle \begin{array}{l}x=\frac{{4x}}{5}+\frac{4}{5}\times 10\\x=\frac{{4x}}{5}+\frac{8}{1}\end{array}\)

\(\displaystyle x=\frac{{4x+40}}{5}\) (by L.C.M.)

\(\displaystyle 5x=4x+40\)

\(\displaystyle 5x-4x=40\) (by transposing)

x = 40 answer

**8. \(\displaystyle \frac{{2x}}{3}+1=\frac{{7x}}{{15}}+3\)**

Sol.: \(\displaystyle \frac{{2x}}{3}+1=\frac{{7x}}{{15}}+3\)

\(\displaystyle \frac{{2x}}{3}-\frac{{7x}}{{15}}=3-1\) (by transposing)

\(\displaystyle \frac{{10x-7x}}{{15}}=\frac{2}{1}\) (by cross multiplication)

10x – 7x = 2 ├Ч 15

3x = 30

\(\displaystyle x=\frac{{30}}{3}\)

x = 10 answer

**9. \(\displaystyle 2y+\frac{5}{3}=\frac{{26}}{3}-y\)**

Sol.: \(\displaystyle 2y+\frac{5}{3}=\frac{{26}}{3}-y\)

\(\displaystyle 2y+y=\frac{{26}}{3}-\frac{5}{3}\)┬а

\(\displaystyle \begin{array}{l}3y=\frac{{26-5}}{3}\\3y=\frac{{21}}{3}\end{array}\) (by transposing)

\(\displaystyle y=\frac{7}{3}\) answer

**10. \(\displaystyle 3m=5m-\frac{8}{5}\)**

Sol.: \(\displaystyle 3m=5m-\frac{8}{5}\)

\(\displaystyle 3m-5m=-\frac{8}{5}\) (by transposing)

\(\displaystyle -2m=-\frac{8}{5}\) (both side have – sign so – to – cancel)

\(\displaystyle m=\frac{8}{{2\times 5}}\)

\(\displaystyle m=\frac{4}{5}\) answer